Mixing Problem HELP!

  • Thread starter the7joker7
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  • #1
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Homework Statement



A tank contains `100` kg of salt and `2000` L of water. A solution of a concentration `0.025` kg of salt per liter enters a tank at the rate `6` L/min. The solution is mixed and drains from the tank at the same rate.

Find the equation for the amount of salt in the tank after t hours.

The Attempt at a Solution



I have 0.15kg/min as my rate in and y(t)/333.33 as my rate out.

Which I put together in the form

dy/dt = (5 - y(t))/(333.33)

Split it into

int of (dy/5 - y) = int of (dt/333.33)

-ln(5 - y) = t/333.33 + C

y(0) = 100, so C = -ln(-95)

-ln(5 - y) = t/(333.33) - ln(-95)

5 - y = -95e^(-t/333.33)

y(t) = 5 + 95e^(-t/333.33)

But that's wrong. I know I've asked alot of stuff today, but this one is driving me crazy, I've tried everything I could think of and nothing came out right. Help!
 

Answers and Replies

  • #2
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This involves chemistry friend.

n1v1=n2v2
 
  • #3
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I'm...fairly sure it doesn't, because we've done problems like this before in Calculus without any Chemistry.
 
  • #4
dynamicsolo
Homework Helper
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I have 0.15kg/min as my rate in and y(t)/333.33 as my rate out.
This is correct. I believe you dropped a decimal place, because the next line should be

dy/dt = (50 - y(t))/(333.33) ,

which now gives you the incoming mass rate of 0.15 kg/min. The asymptotic mass of salt in the tank should be 50 kg
( = 0.025 kg/L · 2000 L), which is what the corrected mass function

y(t) = 50 + 50e^(-t/333.33)

will give you.

I use decimals, rather than fractions, which avoided this headache...
 
  • #5
335
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I'm...fairly sure it doesn't, because we've done problems like this before in Calculus without any Chemistry.
I never said that it cannot be solved without using chemistry.
I wanted to say that this question becomes a piece of meat when solved using chemistry.
 

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